# Derive expression for energy loss when two charged capacitors are connected with like plates together

### Capacitance Long Question Solution

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#### Viva Voice Questions # Capacitance - Long Question Answer 4

1. Derive an expression for energy loss when two charged capacitors are connected with like plates together.

When two capacitors charged at different potential differences and are connected by a metallic wire, there is flow of charge from the capacitor at higher potential to the capacitor at lower potential till the potential of the two capacitors becomes equal. In this process, the total charge remains constant but the total energy before joining the capacitors will be greater than that after joining. This means when two charged capacitors are joined by a wire, there is always a loss of energy.

Consider two charged capacitors of capacitances C1 and C2 that store the charges Q1 and Q2 respectively. Let V1 and V2 be the potentials developed on the respective capacitors.

Energy stored in the capacitor C1 before joining = ½ C1V12
And energy stored in the capacitor C1 before joining = ½ C2V22
.: Total energy before joining, E1 = ½ C1V12 + ½ C2V2….... (i)

When the capacitors are connected by a wire, they will share charge till they acquire a common potential V. Then this common potential V is given by;
Common potential (V) = total charge / total capacitance
Or, Common potential (V)= Q1 + Q2 / C1 + C­2
Or, Common potential (V)= C1V1 + C2V2/ C1 + C2  ……. (ii)

Total energy of a capacitors after connection is given by;
E2 = ½ (total capacity) * (common potential)
= ½ (C1 + C2) * (C1V1 + C2V2/ C1 + C2)2
= ½ (C1V1 + C2V2)2/ (C1 + C2)  ……… (iii)

The loss of energy due to connection is,
E1 – E2     = [½ C1V12 + ½ C2V22]– [ ½ (C1V1 + C2V2)2/ (C1 + C2)]
= [(C1 + C2)(C1V12 + C2V22) – (C1V1 + C2V2)2] / 2(C1 + C2
= C1C(V12 + V22 – 2V1V) / 2(C1 + C2)
.: E1 – E2 = C1C2 (V2 – V1)/ 2(C1 + C2)

This is the expression for the loss of energy after joining the charged capacitors.

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